Problem-30

Question

Does the following equation have a solution for \(A\in \mathbb{R}^{2\times 2}\):

\[ \begin{equation*} A^{2} +A+I=0 \end{equation*} \]

Solution

Let \(A=\begin{bmatrix} a & b\\ c & d \end{bmatrix}\), then we have:

\[ \begin{equation*} \begin{aligned} A^{2} +A+I & =\begin{bmatrix} a^{2} +bc+a+1 & ab+bd+b\\ ac+cd+c & bc+d^{2} +d+1 \end{bmatrix}\\ & \\ & =\begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix} \end{aligned} \end{equation*} \]

From this, we get:

\[ \begin{equation*} \begin{aligned} \left( a^{2} +a+1\right) +bc & =0\\ \left( d^{2} +d+1\right) +bc & =0\\ b( a+d+1) & =0\\ c( a+d+1) & =0 \end{aligned} \end{equation*} \]

We know that \(a^{2} +a+1 >0\) for all \(a\in \mathbb{R}\). This shows that \(bc< 0\). This means that one of \(b\) or \(c\) is less than zero. Without loss of generality, let \(b >0\) and \(c< 0\). This means that \(a+d+1=0\). Let \(a=0,d=-1,b=1,c=-1\). Plugging this into \(A\):

\[ \begin{equation*} A=\begin{bmatrix} 0 & 1\\ -1 & -1 \end{bmatrix} \end{equation*} \]

Verify:

\[ \begin{equation*} \begin{aligned} \begin{array}{l} A^{2} +A+I\\ \end{array} & =\begin{bmatrix} 0 & 1\\ -1 & -1 \end{bmatrix}\begin{bmatrix} 0 & 1\\ -1 & -1 \end{bmatrix} +\begin{bmatrix} 0 & 1\\ -1 & -1 \end{bmatrix} +\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\\ & \\ & =\begin{bmatrix} -1 & -1\\ 1 & 0 \end{bmatrix} +\begin{bmatrix} 0 & 1\\ -1 & -1 \end{bmatrix} +\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\\ & \\ & =\begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix} \end{aligned} \end{equation*} \]

Thus the matrix equation \(A^{2} +A+I=0\) has at least one solution. This is similar to this problem. Though the scalar version — \(a^2 + a + 1 = 0\) — has no real solutions, the matrix version — \(A^2 + A + I = 0\) — has at least one solution in \(\mathbb{R}^{2 \times 2}\).